This problem is a Double Reactant. "K+ H20 ----> KOH + H2" would equal "2K + 2H20 ---> 2KOH + H2"
Balanced :2 K + 2 H2O ----> 2 KOH + H2
The balanced chemical equation for the reaction between potassium (K) and sulfuric acid (H2SO4) to form potassium sulfate (K2SO4) and hydrogen gas (H2) is: 2K + H2SO4 -> K2SO4 + H2
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and potassium (K) is: 2HCl + 2K -> 2KCl + H2.
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
You need to have values for K, Br2 and 2KBr in order to find out the balance. For example, if you had the values as H2 + O2 = H20 the balanced chemical equation would be 10 H2 = H20.
k+h2o>koh+h2
Balanced :2 K + 2 H2O ----> 2 KOH + H2
The balanced chemical equation for the reaction between potassium (K) and sulfuric acid (H2SO4) to form potassium sulfate (K2SO4) and hydrogen gas (H2) is: 2K + H2SO4 -> K2SO4 + H2
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and potassium (K) is: 2HCl + 2K -> 2KCl + H2.
Balanced equation: 2 H2 + O2 --> 2 H2O
2 K + Br2 -> 2 Kbr
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
You need to have values for K, Br2 and 2KBr in order to find out the balance. For example, if you had the values as H2 + O2 = H20 the balanced chemical equation would be 10 H2 = H20.
The balanced chemical equation for the reaction of mercury (Hg) with hydrochloric acid (HCl) is: Hg + 2HCl -> HgCl2 + H2
2K(s) + 2H2O(l) = 2KOH(aq) + H2(g) The eq'n is balanced. The coefficients ( molar ratios) in order of the reation eq'n are 2:2:::2:1
Given the reaction 2K + 2H2O -> 2KOH + H2, we can see that 2 moles of potassium react to produce 1 mole of hydrogen gas. At STP, 1 mole of any gas occupies 22.4 L. Therefore, for 1.6 L of hydrogen gas, 2 moles of potassium are needed, which is approximately 155 g of potassium.
Halving the concentration of H2 will decrease the rate of the reaction, assuming it is a first-order reaction with respect to H2. Since the rate law is rate = k[NO]^2[H2], cutting the concentration of H2 in half will decrease the rate of the reaction by a factor of 0.5.